Timeline for Elementary proof that an open subset of $\Bbb{R}^n$ does not have measure zero?
Current License: CC BY-SA 4.0
5 events
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| Apr 5, 2021 at 13:46 | comment | added | Abdelmalek Abdesselam | @DirkWerner: You are right of course, but in my view, such a proof based on the machinery of the Riemann integral would be a bit of a hack. The combinatorial lemma is stated in very elementary terms and I would prefer a proof which is as elementary. | |
| Apr 3, 2021 at 20:45 | comment | added | Dirk Werner | If you are willing to use Riemann integrals, the combinatorial lemma can be derived by integrating the indicator functions of the intervals involved. | |
| Apr 2, 2021 at 22:55 | history | edited | Abdelmalek Abdesselam | CC BY-SA 4.0 |
deleted 33 characters in body
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| Apr 2, 2021 at 20:50 | comment | added | Abdelmalek Abdesselam | In fact no need for double induction. Induction on $p$ is enough as in the nice answer by Iosif. | |
| Apr 2, 2021 at 20:35 | history | answered | Abdelmalek Abdesselam | CC BY-SA 4.0 |